I remember my professor in college challenging me with this question, which I failed to answer satisfactorily: I know there exists a bijection between the rational numbers and the natural numbers, but can anyone produce an explicit formula for such a bijection?

I wasn't familiar with pairing functions, so let me look at that more closely. My professor insisted, though, that I come up with a formula, and of course that would also require that equivalent pairs (in the rational number sense) shouldn't get counted more than once.
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Alex BassonOct 24 '10 at 4:55

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@lhf. Maybe you should post your comment as an answer; otherwise, it's not unlikey that this question remains unanswered.
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a.r.Oct 24 '10 at 5:21

Could you provide a list of features that you consider legitimate to include in your formula? Often when these questions are posed, responses are met with "that doesn't count as a formula."
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Douglas S. StonesOct 24 '10 at 5:43

Am I missing something or is the Cantor pairing function is not a bijection between the rationals and the naturals? It seems to, at best, reduce our problem to finding a bijection between $\mathbb{Q}$ and $\mathbb{N} \times \mathbb{N}$
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nhamApr 30 '13 at 5:14

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Indeed, none of these are bijections between $\mathbb N$ and $\mathbb Q$, at least, not in the sense of "explicit."
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Thomas AndrewsDec 14 '14 at 16:31

(Defining for positive $n$ the function $f(n)=\rho(n-1)$, we actually have that $f$ is multiplicative - $f(mn)=f(m)f(n)$ of $m,n$ relatively prime. So we get that $f(p^{2k})=p^k$ and $f(p^{2k-1})=p^{-k}$.)

Basically, every positive integer has a unique representation as:

$$p_1^{a_1}p_2^{a_2}\dots$$

where the $p_i$ are all primes, the $a_i$ are non-negative integers, and all but finitely many of the $a_i$ are zero.

The natural numbers have the same representation, with the only change that the $a_i$ are integers, possibly non-negative.

For your bijection between $\mathbb{N}$ and $\mathbb{Q}$, why have you defined it as $\rho_1$ instead of just $\eta$ or something else (I don't see anything to indicate significance of the index)?
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fancynancyFeb 16 at 17:52

@fancynancy I think just because it was a variant of $\rho$. I think the function I now call $\rho_1$ was just called $\rho$, but then I realized I needed the intermediate function, which was more "important" in some way, so I called that $\rho$ and renamed this $\rho_1$. It's just a function name.
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Thomas AndrewsFeb 16 at 18:13

That's what I thought--just wanted to make sure. Thanks for clarifying.
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fancynancyFeb 16 at 18:17

Since $f$ is a bijection this implies that $f^{-1}$ exists. That means given a rational number we can find the corresponding natural number. For example suppose you have a fraction, say it is $1/4$. Can we determine the $n$ such that $f(n)=1/4$? The answer is a resounding yes. Given a positive rational number, $q \in \mathbb{Q}$, the $n$ such that $f(n)=q$ is found by $n=f^{-1}(q)$. This function, $f^{-1}$, is given as follows:

As an example, we see from above that $f(5)={3/2}$. Let us plug $(3/2)$ into $f^{-1}$ and see if we get 5.

$f^{-1}(3/2)=2f^{-1} \bigg(\dfrac{3/2}{1-(3/2)} \bigg)+1=2f^{-1} \bigg(\dfrac{1}{2} \bigg)+1.$ A quick calculation yields that $f^{-1} \bigg(\dfrac{1}{2} \bigg)=2$ and so we get $f^{-1}(3/2)=2f^{-1} \bigg(\dfrac{1}{2} \bigg)+1=2(2)+1=5$.

Step Two: We showed there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}^{+}$. We now attempt to show there exists an explicit bijection between $\mathbb{N}$ and $\mathbb{Q}$. Using the work done in Step One, it appears easier to first create a bijection between $\mathbb{Z}$ and $\mathbb{Q}$. The reason for doing so is because we have already created a bijection from the positive integers (natural numbers) to the positive rationals. So it only seems natural that by adding in the negative integers, we can map them to the negative rationals and thus obtain a bijection. We do this as follows:

We already referenced a proof by Northshield showing that $g(z)=\dfrac{a_{z}}{a_{z+1}}$ if $z>0$ is a bijection from $\mathbb{N} \rightarrow \mathbb{Q}^{+}$. Equivalently, we may write this as $g$ is a bijection from $\mathbb{Z}^{+}$ to $\mathbb{Q}^{+}$ for $z>0$. Now, it follows by the symmetry of the problem that $g(z)=- \dfrac{a_{-z}}{a_{-(z-1)}}$ is a bijection from $\mathbb{Z}^{-}$ to $\mathbb{Q}^{-}$ if $z<0$. That is, $g$ is a bijection between the negative integers and the negative rationals. So we have covered all the positive and negative rationals. The only element in the rationals that is not accounted for is the zero element. So we shall have the integer $0$ mapping to the rational number $0$. However, $g$ is a bijection from the integers to the rationals. We wish to find a bijection from the natural numbers to the rationals. So we shall now define the well-known bijection from the natural numbers to the integers.

You may check for yourself that $h$ is a bijection. It follows that $g~\circ~ h: \mathbb{N} \rightarrow \mathbb{Q}$ is a bijection since the composition of two bijections is a bijection. Thus, we have an explicit bijection from $\mathbb{N}$ to $\mathbb{Q}$.

However, given a rational number, can we find what this rational number maps to in the set of natural numbers? Although I do not prove it, the answer is yes and is given by the following piece-wise defined function which is an extension of the function defined in Step One. We first define $g^{-1}: \mathbb{Q} \rightarrow \mathbb{Z}$ as